import java.util.*;

public class BinaryTree {
 
    static class TreeNode {
        public char val;
        public TreeNode left;//左孩子的引用
        public TreeNode right;//右孩子的引用
 
        public TreeNode(char val) {
            this.val = val;
        }
    }
 
 
    /**
     * 创建一棵二叉树 返回这棵树的根节点
     *
     * @return
     */
    public TreeNode createTree() {
        TreeNode A = new TreeNode('A');
        TreeNode B = new TreeNode('B');
        TreeNode C = new TreeNode('C');
        TreeNode D = new TreeNode('D');
        TreeNode E = new TreeNode('E');
        TreeNode F = new TreeNode('F');
        TreeNode G = new TreeNode('G');
        TreeNode H = new TreeNode('H');

        A.left = B;
        A.right = C;
        B.left = D;
        B.right = E;
        C.left = F;
        C.right = G;
        E.right = H;

        return A;
    }
 
    // 前序遍历
    public void preOrder(TreeNode root) {
        if (root == null) {
            return;
        }

        System.out.print(root.val + " ");
        preOrder(root.left);
        preOrder(root.right);
    }
 
    // 中序遍历
    void inOrder(TreeNode root) {
        if (root == null) {
            return;
        }

        inOrder(root.left);
        System.out.print(root.val +" ");
        inOrder(root.right);

    }
 
    // 后序遍历
    void postOrder(TreeNode root) {
        if (root == null) {
            return;
        }

        postOrder(root.left);
        preOrder(root.right);
        System.out.print(root.val + " ");
    }

    /**
     * 获取树中节点的个数：遍历思路
     */
    public static int nodeSize = 0;
    int size(TreeNode root) {
        if (root == null) {
            return 0;
        }
        nodeSize++;
        size(root.left);
        size(root.right);
        return nodeSize;
    }
 
    /**
     * 获取节点的个数：子问题的思路
     *
     * @param root
     * @return
     */

    int size2(TreeNode root) {
        if (root == null) {
            return 0;
        }
        return size2(root.left) + size2(root.right)+1;
    }
 
 
    /*
     获取叶子节点的个数：遍历思路
     */
    public static int leafSize = 0;
 
    void getLeafNodeCount1(TreeNode root) {
        if (root == null) {
            return;
        }

        if (root.left == null && root.right == null) {
            leafSize++;
        }
        getLeafNodeCount1(root.left);
        getLeafNodeCount1(root.right);
    }
 
    /*
     获取叶子节点的个数：子问题
     */
    int getLeafNodeCount2(TreeNode root) {
        if (root == null) {
            return 0;
        }

        if (root.left == null && root.right == null) {
            return 1;
        }
        return getLeafNodeCount2(root.left) + getLeafNodeCount2(root.right);
    }
 
    /*
    获取第K层节点的个数
     */
    int getKLevelNodeCount(TreeNode root, int k) {
        if (root == null || k<=0) {
            return 0;
        }
        if (k == 1) {
            return 1;
        }
        return getKLevelNodeCount(root.left, k-1) + getKLevelNodeCount(root.right, k-1);
    }
 
    /*
     获取二叉树的高度
     时间复杂度：O(N)
     */
    int getHeight(TreeNode root) {
        if (root == null) {
            return 0;
        }
        int leftHeight = getHeight(root.left);
        int rightHeight = getHeight(root.right);
        return leftHeight > rightHeight ? leftHeight+1 : rightHeight+1;
    }
 
 
    // 检测值为value的元素是否存在
    TreeNode find(TreeNode root, char val) {
        if (root == null) {
            return null;
        }
        if (root.val == val) {
            return root;
        }
        TreeNode ret1 = find(root.left, val);
        if (ret1 != null) {
            return ret1;
        }
        TreeNode ret2 = find(root.right, val);
        if (ret2 != null) {
            return ret2;
        }
        return null;
    }
 
    //层序遍历
    void levelOrder(TreeNode root) {
        if (root == null) {
            return;
        }
        Queue<TreeNode> qu = new LinkedList<>();
        qu.offer(root);
        while (!qu.isEmpty()) {
            TreeNode cur = qu.poll();
            System.out.print(cur.val + " ");
            if (cur.left != null) {
                qu.offer(cur.left);
            }
            if (cur.right != null) {
                qu.offer(cur.right);
            }
        }
    }

    // 二叉树的最大深度
    public int maxDepth(TreeNode root) {
        if (root == null) {
            return 0;
        }
        int leftDepth = maxDepth(root.left);
        int rightDepth = maxDepth(root.right);
        return leftDepth > rightDepth ? leftDepth + 1 : rightDepth + 1;
    }
    // 判断是不是平衡二叉树
    public boolean isBalanced(TreeNode root) {
        if (root == null) {
            return true;
        }
        int l = maxDepth(root.left);
        int r = maxDepth(root.right);
        return Math.abs(l - r) < 2 && isBalanced(root.left) && isBalanced(root.right);
    }

    public int maxDepth2(TreeNode root) {
        if (root == null) {
            return 0;
        }
        int leftDepth = maxDepth2(root.left);
        int rightDepth = maxDepth2(root.right);
        if (leftDepth >= 0 && rightDepth >= 0 && Math.abs(leftDepth-rightDepth) <= 1) {
            return Math.max(leftDepth, rightDepth) + 1;
        } else {
            return -1;
        }
    }
    public boolean isBalanced2(TreeNode root) {
        if (root == null) {
            return true;
        }
        return maxDepth2(root) >= 0;
    }
    // https://leetcode.cn/problems/subtree-of-another-tree/submissions/
    public boolean isSubtree(TreeNode root, TreeNode subRoot) {
        if (root==null && subRoot!=null || root!=null && subRoot==null) {
            return false;
        }
        if (root==null || subRoot==null) {
            return false;
        }
        //1、判断两棵树 是不是两颗相同的树
        if (isSameTree(root, subRoot)) {
            return true;
        }
        //2、subRoot是不是 root.left 子树
        if (isSubtree(root.left, subRoot)) {
            return true;
        }
        //3、subRoot是不是 root.right 子树
        return isSubtree(root.right, subRoot);
    }
    public boolean isSameTree(TreeNode p, TreeNode q) {
        if(p == null && q != null  || p != null && q == null ) {
            return false;
        }
        if(p == null && q == null) {
            return true;
        }
        if(p.val != q.val) {
            return false;
        }
        return isSameTree(p.left,q.left) && isSameTree(p.right,q.right);
    }

    // 判断一棵树是不是完全二叉树
    boolean isCompleteTree(TreeNode root) {
        if (root == null) {
            return true;
        }
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        while (!queue.isEmpty()) {
            TreeNode cur = queue.poll();
            if (cur != null) {
                queue.offer(cur.left);
                queue.offer(cur.right);
            } else {
                break;
            }
        }
        // 这时判断队列里面的元素是不是全是空元素
        while (!queue.isEmpty()) {
            TreeNode cur = queue.poll();
            if (cur != null) {
                return false;
            }
        }
        return true;
    }

    // https://leetcode.cn/problems/lowest-common-ancestor-of-a-binary-tree/
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        Stack<TreeNode> stack1 = new Stack<>();
        Stack<TreeNode> stack2 = new Stack<>();
        getPath(root, p, stack1);
        getPath(root, q, stack2);
        int sz1 = stack1.size();
        int sz2 = stack2.size();
        if (sz1 > sz2) {
            while (sz1 != sz2) {
                stack1.pop();
                sz1--;
            }
        } else if (sz2 > sz1) {
            while (sz2 != sz1) {
                stack2.pop();
                sz2--;
            }
        }
        while (!stack1.isEmpty()) {
            TreeNode node = stack1.pop();
            if (node == stack2.pop()) {
                return node;
            }
        }
        return null;
    }
    public boolean getPath(TreeNode root, TreeNode get, Stack<TreeNode> stack) {
        if (root == null) {
            return false;
        }
        stack.push(root);
        if (root == get) {
            return true;
        }
        // 判断树的左边
        boolean flg1 = getPath(root.left, get, stack);
        if (flg1) {
            return true;
        }
        // 树的左边没有 判断树的右边
        boolean flg2 = getPath(root.right, get, stack);
        if (flg2) {
            return true;
        }
        stack.pop();
        // 树的左右都没有 返回false
        return false;
    }
    // https://leetcode.cn/problems/binary-tree-level-order-traversal-ii/
    public List<List<Character>> levelOrder2(TreeNode root) {
        List<List<Character>> lists = new ArrayList<>();
        if (root == null) {
            return lists;
        }
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        while (!queue.isEmpty()) {
            int sz = queue.size();
            List<Character> list = new ArrayList<>();
            while (sz > 0) {
                TreeNode cur = queue.poll();
                sz--;
                list.add(cur.val);
                if (cur.left != null) {
                    queue.offer(cur.left);
                }
                if (cur.right != null) {
                    queue.offer(cur.right);
                }
            }
            lists.add(list);
        }
        return lists;
    }

    // https://leetcode.cn/problems/binary-tree-level-order-traversal-ii/
    public List<List<Character>> levelOrderBottom(TreeNode root) {
        List<List<Character>> lists = new ArrayList<>();
        if (root == null) {
            return lists;
        }
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        while (!queue.isEmpty()) {
            int sz = queue.size();
            List<Character> list = new ArrayList<>();
            while (sz > 0) {
                TreeNode cur = queue.poll();
                sz--;
                list.add(cur.val);
                if (cur.left != null) {
                    queue.offer(cur.left);
                }
                if (cur.right != null) {
                    queue.offer(cur.right);
                }
            }
            lists.add(list);
        }
        Collections.reverse(lists);
        return lists;
    }
}